Question 201614: How do you know if a quadratic equation will have one, two, or no solutions? How do you find a quadratic equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? Explain. Provide your classmate’s with one or two solutions with which they must create a quadratic equation
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Answer by jsmallt9(3758)   (Show Source):
Answer by solver91311(24713)  (Show Source):
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How do you know if a quadratic equation will have one, two, or no solutions?.
The short answer is, it is easy: They all have two solutions. The Fundamental Theorem of Algebra guarantees it. Now if what you really meant to ask is: How do you know if a quadratic equation will have one, two, or no solutions over the real numbers, then read on.
A quadratic equation of the form:
has a general solution:
The expression under the radical in the general solution, namely  is called the discriminant. The discriminant can be evaluated to determine the character of the solutions of a quadratic equation, thus:
If  , then
if  , then the quadratic has two distinct real number roots. Furthermore, if  is a perfect square number, then the roots will be rational, otherwise the roots of the equation will be a conjugate pair of irrational numbers of the form  where 
if  , then the quadratic has a single real number root with a multiplicity of 2. In this case the quadratic is a perfect square having two factors: (x - \alpha)) , hence  is the root, and the multiplicity of 2 comes from the fact that there are two identical factors.
if  , then the quadratic has no real number solutions. It has a conjugate pair of complex roots of the form  where  and  is the imaginary number defined by 
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How do you find a quadratic equation if you are only given the solution?
Note, that should read "solutions"
If the solutions of a quadratic equation are and  , then you can say:
 and
 , therefore
(x - \beta) = 0) and finally,
x + \alpha\beta = 0)
Hence deriving the quadratic equation from the solutions. Note that if you are only given one number as a solution, then you have three possibilities. The first is that you may have a rational number, in which case you have two identical factors that become a perfect square trinomial: (x - \alpha) = 0 \ \ \Rightarrow\ \ x^2 - 2\alpha x + \alpha^2 = 0) . The second possibility is that you are given an irrational number, such as  with the possibility that  . In this case, the second root is the conjugate of the given root, namely  . The third possibility is that you are given a complex number root,  . Again, the second root is the conjugate: 
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Is it possible to have different quadratic equations with the same solution?
Well, that depends on what you mean by 'different'. For a given  ,  , and  ,  is equivalent to, and has the same roots as  for all  (actually it is true for all  but let's stick with rational coefficients for your purposes). Since the equations are equivalent, it is a stretch, for me anyway, to call them 'different.' However, what is true is that the analogous functions, i.e.  = k_iax^2 + k_ibx + k_ic) are very different in that they all have different graphs.
So while it is true that  = x^2 - 6x + 5) is different than  = 2x^2 - 12x + 10) (see graphs below), the equations  and  are equivalent because you can multiply the second one by  to obtain the first. Note that the graphs intersect the  -axis at the same points:
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There should be enough information above for you to come up with a set of solutions for a quadratic for your classmates, particularly in the answer to the second question.
John
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